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Problem: If $\sum a_n$ converges, and $\{b_n\}$ is monotonic and bounded, prove that $\sum a_n b_n$ converges.

Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 8.

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1 Answer

up vote 5 down vote accepted

The sequence $\{b_n\}$ is monotonic and bounded, so it converges to some number $C$. Assume, without loss of generality, that the sequence $\{b_n\}$ is increasing, and write $b_n=C-d_n$, where $d_n\rightarrow 0$. We have

$$\sum a_nb_n = C\sum a_n -\sum a_nd_n.$$

The first series on the right is convergent by hypothesis, and the second is convergent because of the following theorem:

Theorem: If the partial sums of $\sum t_n$ form a bounded sequence and $s_n$ is a decreasing sequence that tends to 0, then $\sum t_ns_n$ converges.

Here we take $t_n=a_n$ and $s_n=d_n$.

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Did you just answer your own question! =) –  sidht Jun 26 '12 at 20:56
Since this is a rather old problem so I don't know if you are still around, @Potato: but could you tell me if this approach you did for this problem, especially the part $b_n = C - d_n$ a rather common technique to use ? –  hyg17 Jun 5 '13 at 8:49
@hyg17 I don't think so, but it's been a while since I've done a lot of series manipulation. –  Potato Jun 5 '13 at 13:02
Hi Potato, the theorem you mentioned has been just as difficult for me to prove as the original problem. For example, we have to worry about $\sum t_n$ diverging just as much as we have to worry about $\sum b_n$ diverging. Is there a simple proof to the result you mentioned? –  Dan Douglas Aug 24 '13 at 6:07
@DanDouglas It's slightly tricky. You need to use summation by parts. The following 2 images are excerpts from Rudin's book, pages 70 and 71. Click here. Let me know if they answer your question or if you need more detail. –  Potato Aug 24 '13 at 6:40
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